Question

How many numbers can be made with the help of the digits \[0,1,2,3,4,5\]which are greater than \[3000\](repetition is not allowed)?
A. \[180\]
B. \[360\]
C. \[1380\]
D. \[1500\]

Answer 1

Hint: We will use the concept of permutation for solving the question which states that a permutation is an arrangement of items in a certain direction or sequence. One should consider both the selection and the arrangement while dealing with permutation. In permutations, ordering is crucially important. The permutation is seen as an ordered combination, in other words. The positive integers can either consist of 4 or 5 or 6 digits. Since 3000 is a 4-digit number, so we can get 4 digits numbers which are greater than 3000. It can be noted that any positive integer consisting of 5 or 6 digits will be greater than 3000 therefore, we will find the find the total number of 4 digits, 5 digits & 6 digits numbers formed by permutation and then will add all the numbers to get the answer.

Formula used:
$n! = n(n-1)(n-2)....1 $

Complete step by step solution:
We have total 6 digits here: 0,1,2,3,4,5
More than 3000 can be written as 4-digit, 5-digit, or 6-digit numbers.

The digit of the thousand place of four digits numbers that are greater than 3000 will be either start with 3 or 4 or 5.
So the possible number in thousand is 3.
Since we have only 6 numbers by using that we can make a number.
The repetition of numbers is not allowed, so we have 5 numbers to fill the hundred place of the 4 digit number.
Now we have only 4 numbers by using which we can fill the tenth place of the number and 3 numbers to fill the unit place of the number.
Multiply all possible numbers to find the number of 4-digit numbers.
Thus,
We will first find the total numbers of four-digit numbers formed:
Therefore, four-digit numbers formed \[ = 3 \times 5 \times 4 \times 3\]
\[ = 180\]
For 5-digit numbers:
If a 5-digit number starts with 0 then it becomes 4 digit number.
So we have 1,2,3,4 and 5 to fill the ten thousand place.
The possible number of digits to fill the ten thousand place is 5.
Since we have only 6 numbers and repetition of numbers is not allowed.
So, we have 5 numbers to fill the thousand place of the number after filling ten thousand place.
Similarly, we have 4 numbers to fill the hundred place, 3 numbers to fill ten place, and 2 numbers to fill the unit place of the 5-digit number.
Multiply all possible numbers to find the number of 5-digit numbers.
Now we will find the total of five-digit numbers formed:
Therefore, five-digit numbers formed \[ = 5 \times 5 \times 4 \times 3 \times 2\]
 \[ = 600\]
For 6-digit numbers:
If a 6-digit number starts with 0 then it becomes 5 digit number.
So we have 1,2,3,4 and 5 to fill the hundred thousand place.
The possible number of digits to fill the hundred thousand place is 5.
Since we have only 6 numbers and repetition of numbers is not allowed.
So, we have 5 numbers to fill the ten thousand place of the number after filling a hundred thousand place.
Similarly, we have 4 numbers to fill the thousand place of the 6-digit number, 3 numbers to fill the hundred place of the 6-digit number, 2 numbers to fill the ten place of the 6-digit number, and 1 number to fill the unit place of the 6-digit number.
Multiply all possible numbers to find the number of 6-digit numbers.
Now we will find the total of six-digit numbers formed:
Therefore, six-digit numbers formed \[ = 5 \times 5 \times 4 \times 3 \times 2 \times 1\]
\[ = 600\]
Now we will add total numbers formed:
therefore, total numbers formed \[ = 180 + 600 + 600\]
  \[ = 1380\]
Hence, option C is correct.

Note: Students can make mistakes by applying the factorial formula. Since it is a question related to non-repetition and order matter. Since we are unable to use all given numbers to fill the first place of a number. Hence we are unable to use the factorial formula to calculate the possible numbers.
Here we used the permutation formula to calculate the number of numbers that are made by the given numbers.